1. Key Concepts and Formulas

To accurately solve problems involving corrected statistical measures, a solid understanding of the fundamental definitions and computational formulas for mean and variance is essential.

• Mean ($\overline{x}$): The average of a dataset, calculated by summing all observations and dividing by the number of observations ($N$). $\overline{x} = \frac{\sum x_i}{N}$
• Variance ($\sigma^2$): A measure of data dispersion, indicating how spread out the numbers are from the mean. The computational formula is particularly useful for correction problems as it directly involves the sum of squares: $\sigma^2 = \frac{\sum x_i^2}{N} - (\overline{x})^2$
• Standard Deviation ($\sigma$): The square root of the variance, providing a measure of spread in the same units as the original data. $\sigma = \sqrt{\sigma^2}$

2. Step-by-Step Solution

We are given $N=20$, incorrect mean $\overline{x}_{\text{inc}} = 10$, incorrect standard deviation $\sigma_{\text{inc}} = 2.5$. A value of $25$ was mistakenly taken instead of $35$. We need to find the correct mean $\alpha$ and correct variance $\beta$ (since $\sqrt{\beta}$ is the correct standard deviation).

Step 1: Calculate the Incorrect Sum of Observations ($\sum x_{i, \text{inc}}$)

• What we're doing: We first determine the total sum of all observations as it was originally calculated, including the erroneous value. This is a prerequisite for correcting the sum.
• Why we're doing it: The mean formula directly relates the sum of observations to the mean and the number of observations. We use the given incorrect mean to find this initial sum.
• Using the mean formula: $\overline{x}_{\text{inc}} = \frac{\sum x_{i, \text{inc}}}{N}$
• Substitute the given values: $10 = \frac{\sum x_{i, \text{inc}}}{20}$
• Solve for the incorrect sum: $\sum x_{i, \text{inc}} = 10 \times 20 = 200$

Step 2: Calculate the Correct Sum of Observations ($\sum x_{i, \text{corr}}$)

• What we're doing: We adjust the incorrect sum by removing the contribution of the wrong data point and adding the contribution of the correct data point.
• Why we're doing it: This step directly corrects the cumulative sum of values, which will then allow us to calculate the true mean.
• The correction formula is: $\sum x_{i, \text{corr}} = \sum x_{i, \text{inc}} - (\text{wrong value}) + (\text{correct value})$
• Substitute the values: $\sum x_{i, \text{corr}} = 200 - 25 + 35$ $\sum x_{i, \text{corr}} = 175 + 35 = 210$

Step 3: Calculate the Correct Mean ($\alpha$)

• What we're doing: We compute the correct mean using the newly found correct sum of observations.
• Why we're doing it: This is the primary goal for the mean calculation, using the fundamental definition of mean with the corrected data.
• Using the mean formula with the correct sum: $\alpha = \overline{x}_{\text{corr}} = \frac{\sum x_{i, \text{corr}}}{N}$
• Substitute the values: $\alpha = \frac{210}{20}$ $\alpha = 10.5$ Thus, the correct mean $\alpha$ is $10.5$.

Step 4: Calculate the Incorrect Variance ($\sigma_{\text{inc}}^2$)

• What we're doing: We convert the given incorrect standard deviation into incorrect variance.
• Why we're doing it: The variance formula is used to work with sums of squares, so having the variance directly simplifies subsequent calculations.
• From the given incorrect standard deviation: $\sigma_{\text{inc}}^2 = (2.5)^2 = 6.25$

Step 5: Calculate the Incorrect Sum of Squares of Observations ($\sum x_{i, \text{inc}}^2$)

• What we're doing: We determine the total sum of the squares of all observations as it was originally calculated, including the square of the erroneous value.
• Why we're doing it: To find the correct variance, we need the correct sum of squares. We use the incorrect variance and incorrect mean to work backward and find the initial sum of squares.
• Rearranging the variance formula: $\sigma_{\text{inc}}^2 = \frac{\sum x_{i, \text{inc}}^2}{N} - (\overline{x}_{\text{inc}})^2$ $\sum x_{i, \text{inc}}^2 = N \left( \sigma_{\text{inc}}^2 + (\overline{x}_{\text{inc}})^2 \right)$
• Substitute the known incorrect values: $\sum x_{i, \text{inc}}^2 = 20 \left( 6.25 + (10)^2 \right)$ $\sum x_{i, \text{inc}}^2 = 20 \left( 6.25 + 100 \right)$ $\sum x_{i, \text{inc}}^2 = 20 \times 106.25 = 2125$

Step 6: Calculate the Correct Sum of Squares of Observations ($\sum x_{i, \text{corr}}^2$)

• What we're doing: We adjust the incorrect sum of squares by removing the square of the wrong data point and adding the square of the correct data point.
• Why we're doing it: Variance depends on the sum of squares. This correction is crucial to ensure the subsequent variance calculation is based on the true data.
• The correction formula for sum of squares: $\sum x_{i, \text{corr}}^2 = \sum x_{i, \text{inc}}^2 - (\text{wrong value})^2 + (\text{correct value})^2$
• Substitute the values: $\sum x_{i, \text{corr}}^2 = 2125 - (25)^2 + (35)^2$ $\sum x_{i, \text{corr}}^2 = 2125 - 625 + 1225$ $\sum x_{i, \text{corr}}^2 = 1500 + 1225 = 2725$

Step 7: Calculate the Correct Variance ($\beta$)

• What we're doing: We compute the correct variance using the newly found correct sum of squares and the correct mean.
• Why we're doing it: This is the primary goal for the variance calculation. It's critical to use the correct mean ($\alpha$) calculated in Step 3 for this step.
• Using the variance formula with the corrected values: $\beta = \sigma_{\text{corr}}^2 = \frac{\sum x_{i, \text{corr}}^2}{N} - (\overline{x}_{\text{corr}})^2$
• Substitute the correct sum of squares, number of observations, and the correct mean ($\alpha = 10.5$): $\beta = \frac{2725}{20} - (10.5)^2$ $\beta = 136.25 - 110.25$ $\beta = 26$ Thus, the correct variance $\beta$ is $26$.

Step 8: State the Final Corrected Measures

• We have found the correct mean $\alpha = 10.5$ and the correct variance $\beta = 26$.
• Therefore, the pair $(\alpha, \beta)$ is $(10.5, 26)$.

3. Common Mistakes & Tips

• Always Correct the Sums First: The most critical step in these problems is to first correct the sum of observations ($\sum x_i$) and the sum of squares of observations ($\sum x_i^2$). Recalculate mean and variance only after these sums are accurate.
• Use the Corrected Mean for Variance: A frequent error is using the original (incorrect) mean when calculating the corrected variance. Always use the newly calculated correct mean ($\alpha$) in the variance formula for the corrected data.
• Square Values for Sum of Squares: When correcting the sum of squares, remember to subtract the square of the wrong value and add the square of the correct value, i.e., $-(\text{wrong})^2 + (\text{correct})^2$. Do not simply subtract/add the values themselves.

4. Summary

This problem demonstrates a systematic approach to correcting statistical measures when an error in a single data point is identified. The process involves first determining the initial (incorrect) sums of observations and their squares. These sums are then adjusted by removing the contribution of the erroneous value and adding the contribution of the correct value. Finally, the corrected mean and variance are calculated using these adjusted sums, ensuring that the correct mean is used in the variance calculation. Following these steps leads to the correct mean $\alpha = 10.5$ and the correct variance $\beta = 26$.

5. Final Answer The final answer is $\boxed{(10.5, 26)}$ which corresponds to option (D).